Calculate the standard reaction enthalpy for the reaction NO2(g) → NO(g) + O(g) Given: O2(g) → 2 O(g) ∆H◦ = +498.4 kJ/mol NO(g) + O3(g) → NO2(g) + O2(g) ∆H◦ = −200 kJ/mol 3O2(g) → 2 O3(g) ∆H◦ = +285.4 kJ/mol

∆H◦ = 306.5 kJ

Explanation:

It is possible to obtain standard reaction enthalpy of a reaction by the sum of half-reactions associated. Having:

(1) O2(g) → 2 O(g) ∆H◦ = +498.4 kJ/mol

(2) NO(g) + O3(g) → NO2(g) + O2(g) ∆H◦ = −200 kJ/mol

(3) 3O2(g) → 2 O3(g) ∆H◦ = +285.4 kJ/mol

The rest of 1/2 (1) - (2) gives:

NO2(g) + 3/2 O2(g)  →  O(g) + NO(g) + O3(g)

∆H◦ = 1/2 498.4 kJ/mol - (-200 kJ/mol) = 449.2 kJ  

Now, if you rest 1/2 (3):

NO2(g) → NO(g) + O(g) ∆H◦ = 306.5 kJ

the dependant variable is spilled water

the independent is the heat

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